Motivation:
Let $\mathbb{D}^2$ be the closed unit disk. I am studying the "elastic energy" functional $E(f)=\int_{\mathbb{D}^2} \text{dist}^2(df,\text{SO}_2)$, where $f \in C^{\infty}(\mathbb{D}^2,\mathbb{R}^2)$.
Denoting by $Q(df)$ the closest special orthogonal matrix to $df$, we have $E(f)=\int_{\mathbb{D}^2} |df-Q(df)|^2$, so the behaviour of $Q(df)$ naturally plays a key role in the analysis;
The question:
Let $f:\mathbb{D}^2 \to \mathbb{R}^2$ be real-analytic in the interior of $\mathbb{D}^2$ and smooth on whole $\mathbb{D}^2$. Suppose that $\det df>0$ almost everywhere. Set $$U:= \{ p \in \mathbb{D}^2 \, | \, \det df>0\},$$
and let $Q=Q(df)=df(\sqrt{df^Tdf})^{-1}$ be the orthogonal polar factor of $df$; $\sqrt{}$ is the unique symmetric positive-definite square root. $Q$ is well-defined on $U$-in particular it is defined a.e. on $\mathbb{D}^2$, and the restriction $Q|_U$ is smooth (even real-analytic in $U \cap \text{int}(\mathbb{D}^2)$).
Question: Does $Q \in W^{1,p}_{loc}(\mathbb{D}^2,\mathbb{R}^{4})$ for some $p \ge 1$?
Edit: For a start, I would be very happy to know if the classical derivatives of $Q$ (which exists on $U$) are in $L^1_{loc}(\mathbb{D}^2)$; Even deciding this question of integrability seems non-trivial.
I am not even sure what happens when $f$ is conformal a.e. (so it's actually holomorphic); In that case, $Q(df)=\frac{\sqrt 2}{\|df\|}df$ is obtained from $df$ simply by normalization. I asked about this case separately here.
Note that the Hausdorff dimension of the singular set $\mathbb{D}^2 \setminus U=(\det df)^{-1}(0)$ is not greater than 1. (since it is the zero set of the non-zero real-analytic function $\det df$).
Further discussion:
The only possible problem is when we approach points where $df=0$. Indeed, the polar factor map $\text{GL}_2^+ \to \text{SO}_2$ can be extended to a smooth map from $\text{GL}_2^+ \cup (\text{rank} = 1) \to \text{SO}_2$. (In general dimension $n$, there is a smooth extension to $\text{GL}_n^+ \cup (\text{rank} = n-1)$).
However, the polar factor cannot be chosen continuously on all $\text{GL}_2^+ \cup \det^{-1}(0)$; there are singularities:
$\lim_{t \to 0} Q(\left(\begin{matrix}t & 0 \\ 0 & t\end{matrix}\right))=\text{sgn}(t) \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$, and
$\lim_{t \to 0}Q(\left(\begin{matrix}0 & -t \\ t & 0\end{matrix}\right))=\text{sgn}(t) \left(\begin{matrix}0 & -1 \\ 1 & 0\end{matrix}\right)$.
Since the sign function is not Sobolev, I tried lifting these singularities to create a singularity in $Q=Q(df)$ for some map $f$. My first attempt was to use only one of the two singularities above, but this failed- if $df$ is always of the form $\left(\begin{matrix}t(x,y) & 0 \\ 0 & t(x,y) \end{matrix}\right)$ or always of the form $ \left(\begin{matrix}0 & -t(x,y) \\ t(x,y) & 0\end{matrix}\right)$, then it must be constant. Thus, it seems that in order to create a singularity, we somehow need to create a situation where $df$ interpolates between these two forms.
Here is an "interpolating example", that is still Sobolev (so it does not decide the question):
Set $f(x,y)=(x^2-y^2,2xy)$. Then $df=2\left(\begin{matrix}x & -y \\ y & x\end{matrix}\right)$, so $Q=Q(df)=\left(\begin{matrix}\frac{x}{\sqrt{x^2+y^2}} & -\frac{y}{\sqrt{x^2+y^2}} \\ \frac{y}{\sqrt{x^2+y^2}} & \frac{x}{\sqrt{x^2+y^2}}\end{matrix}\right) \in W^{1,p}$ for every $p\in[1, 2)$. (This example is the holomorphic function $f(z)=z^2$; see here for a further discussion of the holomorphic case).
Final comment: If we allow $df$ to switch orientations, then $Q(df)$ is not always Sobolev. Even in dimension $1$, for $f(x)=x^2$, we have $Q(x)=\text{sgn}(f'(x))=\text{sgn}(x) $ which is not in $ W^{1,1}(\mathbb R,\mathbb R)$. The purpose of this question is to determine if non-negligible singularities may occur when the map $f$ is a.e. orientation-preserving.
(Note that $Q(\left(\begin{matrix}t & 0 \\ 0 & s\end{matrix}\right))= \left(\begin{matrix}\text{sgn}(t) & 0 \\ 0 & \text{sgn}(s)\end{matrix}\right)$, but if we want $\det \left(\begin{matrix}t & 0 \\ 0 & s\end{matrix}\right)=ts \ge 0$, we must force $s,t$ to switch signs "at the same times" (points), i.e. they should be "coupled" somehow, which explains why I chose to take $s=t$ in the examples above).
